Solving Trig/Algebra Questions: Finding the Missing Factor -√2/2

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The discussion centers on a user struggling to simplify a trigonometric expression to arrive at -√2 instead of -√2/2. The expression involves operations with cosine and sine values, leading to confusion about the missing factor. After some back-and-forth, the user realizes that rearranging the terms and multiplying by √2 helps clarify the simplification process. Ultimately, the correct simplification is confirmed as -√2. This exchange highlights the importance of careful manipulation of algebraic expressions in trigonometry.
majinkenji
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Hello, All,

I am unclear what my oversight is on this one -- supposed to be getting -√2 as the answer on these practice worksheets, however I keep getting -√2/2. Where is the factor? Thank you for your guidance

(cos^2 ∏) / (sin 5∏/4)

(-1)^2 / (-√2/2)

(1) * (-2/√2)

(-2/√2) * (√2/√2)

(-√2/2)

?

-mk
 
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In simplifying this, what is the final numerator: (-2/√2) * (√2/√2)
 
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majinkenji said:
(-2/√2) * (√2/√2)

Try re-arranging this as

-2/(√2 √2) * (√2)
 
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got it, thank you very much :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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